Find the set of all values of a for which the roots of the equation `x^2-2ax + a^2 + a-3=0` are less than 3,

Options

A. `alt2`
B. `2leale3`
C. `3ltale4`
D. `agt4`

Answer

To find the set of all values of \( a \) for which the roots of the equation

\[ x^2 - 2ax + (a^2 + a - 3) = 0 \]

are less than 3, we will follow these steps:

### Step 1: Identify the roots of the quadratic equation

The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation, we have:

- \( a = 1 \)
- \( b = -2a \)
- \( c = a^2 + a - 3 \)

Thus, the roots are:

\[ x = \frac{2a \pm \sqrt{(-2a)^2 - 4 \cdot 1 \cdot (a^2 + a - 3)}}{2 \cdot 1} \]

### Step 2: Simplify the expression for the roots

Calculating the discriminant:

\[ b^2 - 4ac = 4a^2 - 4(a^2 + a - 3) = 4a^2 - 4a^2 - 4a + 12 = -4a + 12 \]

Thus, the roots become:

\[ x = \frac{2a \pm \sqrt{-4a + 12}}{2} = a \pm \sqrt{3 - a} \]

### Step 3: Set up inequalities for the roots to be less than 3

We need both roots \( x_1 \) and \( x_2 \) to be less than 3.

1. For the first root \( x_1 = a + \sqrt{3 - a} < 3 \):

\[ a + \sqrt{3 - a} < 3 \]

Subtract \( a \) from both sides:

\[ \sqrt{3 - a} < 3 - a \]

Square both sides (noting that both sides must be non-negative):

\[ 3 - a < (3 - a)^2 \]

Expanding the right side:

\[ 3 - a < 9 - 6a + a^2 \]

Rearranging gives:

\[ 0 < a^2 - 5a + 6 \]

Factoring:

\[ 0 < (a - 2)(a - 3) \]

This inequality holds when \( a < 2 \) or \( a > 3 \).

2. For the second root \( x_2 = a - \sqrt{3 - a} < 3 \):

\[ a - \sqrt{3 - a} < 3 \]

Subtract \( a \) from both sides:

\[ -\sqrt{3 - a} < 3 - a \]

Multiplying by -1 (reversing the inequality):

\[ \sqrt{3 - a} > a - 3 \]

Square both sides:

\[ 3 - a > (a - 3)^2 \]

Expanding gives:

\[ 3 - a > a^2 - 6a + 9 \]

Rearranging gives:

\[ 0 > a^2 - 5a + 6 \]

Factoring:

\[ 0 > (a - 2)(a - 3) \]

This inequality holds when \( 2 < a < 3 \).

### Step 4: Find the intersection of the two conditions

From the first condition, we have \( a < 2 \) or \( a > 3 \).

From the second condition, we have \( 2 < a < 3 \).

The only overlapping interval is:

\[ a < 2 \]

### Final Answer

Thus, the set of all values of \( a \) for which the roots of the equation are less than 3 is:

\[ \boxed{(-\infty, 2)} \]